Question

Find the matrix A, such that \[\left[ {\begin{array}{*{20}{c}}
  4 \\
  1 \\
  3
\end{array}} \right]A = \left[ {\begin{array}{*{20}{c}}
  { - 4}&8&4 \\
  { - 1}&2&1 \\
  { - 3}&6&3
\end{array}} \right]\].

Answer 1

Hint: We will first assume that the matrix A is \[\left[ {\begin{array}{*{20}{c}}
  x&y&z
\end{array}} \right]\]. Then we will use this value of A in the given equation and simplify. Then we will use that the corresponding elements of two equal matrices are equal to find the required value.

Complete step by step answer:

We are given that
\[\left[ {\begin{array}{*{20}{c}}
  4 \\
  1 \\
  3
\end{array}} \right]A = \left[ {\begin{array}{*{20}{c}}
  { - 4}&8&4 \\
  { - 1}&2&1 \\
  { - 3}&6&3
\end{array}} \right]{\text{ ......eq.(1)}}\]

Let us assume that the matrix A is \[\left[ {\begin{array}{*{20}{c}}
  x&y&z
\end{array}} \right]\].

Using the value of A in the equation (1), we get

\[
   \Rightarrow \left[ {\begin{array}{*{20}{c}}
  4 \\
  1 \\
  3
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  x&y&z
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  { - 4}&8&4 \\
  { - 1}&2&1 \\
  { - 3}&6&3
\end{array}} \right] \\
   \Rightarrow \left[ {\begin{array}{*{20}{c}}
  {4x}&{4y}&{4z} \\
  x&y&z \\
  {3x}&{3y}&{3z}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  { - 4}&8&4 \\
  { - 1}&2&1 \\
  { - 3}&6&3
\end{array}} \right] \\
 \]

Now we know that the corresponding elements of two equal matrices are equal, we have
\[ \Rightarrow 4x = - 4{\text{ ......eq.(2)}}\]
\[ \Rightarrow 4y = 8{\text{ .......eq.(3)}}\]
\[ \Rightarrow 4z = 4{\text{ .......eq.(4)}}\]

Dividing the equation (2) by 4 on both sides, we get
\[
   \Rightarrow \dfrac{{4x}}{4} = \dfrac{{ - 4}}{4} \\
   \Rightarrow x = - 1 \\
 \]

Dividing the equation (3) by 4 on both sides, we get
\[
   \Rightarrow \dfrac{{4y}}{4} = \dfrac{8}{4} \\
   \Rightarrow y = 2 \\
 \]

Dividing the equation (4) by 4 on both sides, we get
\[
   \Rightarrow \dfrac{{4z}}{4} = \dfrac{4}{4} \\
   \Rightarrow z = 1 \\
 \]

Thus, the values are \[x = - 1\], \[y = 2\] and \[z = 1\].

Hence, the matrix A is \[\left[ {\begin{array}{*{20}{c}}
  { - 1}&2&1
\end{array}} \right]\].

Note: In solving these types of questions, you should be familiar with matrices, their addition, and multiplications. The key part of solving this question is by knowing that the corresponding elements of two equal matrices are equal. Avoid calculation mistakes.